Solve x^2-2x-19leq0

$${ x }^{ 2 } -2x-19 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}1-2\sqrt{5},2\sqrt{5}+1\end{bmatrix}$$

Steps Using the Quadratic Formula

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}-2x-19=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -2 for b, and -19 for c in the quadratic formula.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\left(-19\right)}}{2}$$

Do the calculations.

$$x=\frac{2±4\sqrt{5}}{2}$$

Solve the equation x=\frac{2±4\sqrt{5}}{2} when ± is plus and when ± is minus.

$$x=2\sqrt{5}+1 x=1-2\sqrt{5}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(2\sqrt{5}+1\right)\right)\left(x-\left(1-2\sqrt{5}\right)\right)\leq 0$$

For the product to be ≤0, one of the values x-\left(2\sqrt{5}+1\right) and x-\left(1-2\sqrt{5}\right) has to be ≥0 and the other has to be ≤0. Consider the case when x-\left(2\sqrt{5}+1\right)\geq 0 and x-\left(1-2\sqrt{5}\right)\leq 0.

$$x-\left(2\sqrt{5}+1\right)\geq 0 x-\left(1-2\sqrt{5}\right)\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(2\sqrt{5}+1\right)\leq 0 and x-\left(1-2\sqrt{5}\right)\geq 0.

$$x-\left(1-2\sqrt{5}\right)\geq 0 x-\left(2\sqrt{5}+1\right)\leq 0$$

The solution satisfying both inequalities is x\in \left[1-2\sqrt{5},2\sqrt{5}+1\right].

$$x\in \begin{bmatrix}1-2\sqrt{5},2\sqrt{5}+1\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}1-2\sqrt{5},2\sqrt{5}+1\end{bmatrix}$$